Question: Let $f(x, y) = e^{x^2 + y^2}$ and $g(t) = (4t, 3t)$. $h(t) = f(g(t))$ $h'(t) = $
Answer: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (4t, 3t) \\ \\ &g'(t) = (4, 3) \\ \\ &\nabla f = \left( 2xe^{x^2 + y^2}, 2ye^{x^2 + y^2} \right) \\ \\ &\nabla f(g(t)) = \left( 8te^{25t^2}, 6te^{25t^2} \right) \end{aligned}$ Substituting: $\begin{aligned} h'(t) &= \left( 8te^{25t^2}, 6te^{25t^2} \right) \cdot (4, 3) \\ \\ &= 32te^{25t^2} + 18te^{25t^2} \\ \\ &= 50te^{25t^2} \end{aligned}$ Answer $h'(t) = 50te^{25t^2}$